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Integrals

Multiple Choice Questions

671.

The value of $\int \sqrt{1 + \sec (x)} dx$ is

$\sin^{- 1} (\sqrt{2} \sin (x)) + C$
$2 \sin^{- 1} (\sqrt{2} \sin (x / 2)) + C$
$2 \sin^{- 1} (\sqrt{2} \sin (x)) + C$
$2 \sin^{- 1} (\sqrt{2} x / 2) + C$

672.

The value of $\int \frac{(x^{2} + 1)}{x^{4} + x^{2} + 1} dx$ is

$\frac{1}{\sqrt{3}} \tan^{- 1} \{\frac{x - 1 / x}{\sqrt{3}}\} + C$
$\frac{1}{2 \sqrt{3}} \log \{\frac{(x - 1 / x) - \sqrt{3}}{(x - 1 / x) + \sqrt{3}}\} + C$
$\tan^{- 1} \{\frac{x + 1 / x}{\sqrt{3}}\} + C$
$\tan^{- 1} \{\frac{x - 1 / x}{\sqrt{3}}\} + C$

673.

The value of $\int_{0}^{1} x^{2} {(1 - x^{2})}^{\frac{3}{2}} dx$ is

$\frac{1}{32}$
$\frac{π}{8}$
$\frac{π}{16}$
$\frac{π}{32}$

674.
The value of $\int_{0}^{\infty} \frac{x}{(1 + x) (x^{2} + 1)} dx$ is

$2 π$

$\frac{π}{4}$

$\frac{π}{16}$

$\frac{π}{32}$

$\frac{π}{4}$

$Let I = \int_{0}^{\infty} \frac{x}{(1 + x) (x^{2} + 1)} dx By partial fraction, \frac{x}{(1 + x) (x^{2} + 1)} = \frac{A}{(1 + x)} + \frac{Bx + C}{(x^{2} + 1)} \Rightarrow x = A (x^{2} + 1) + (1 + x) (Bx + C) \Rightarrow x = A (x^{2} + 1) + (Bx + {Bx}^{2} + C + Cx) \Rightarrow x = (A + B) x^{2} + (B + C) x + (A + C) On comparing both sides, we get A + B = 0, B + C = 1, A + C = 0 . . . (i) On adding all these equations, we get A + B + C = \frac{1}{2} . . . (ii) ∴ A = \frac{1}{2} - 1 = - \frac{1}{2}, C = \frac{1}{2} and B = \frac{1}{2} ∴ I = \int_{0}^{\infty} \{\frac{- 1}{2 (1 + x)} + \frac{1 (x + 1)}{2 (x^{2} + 1)}\} dx = - \frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x} + \frac{1}{2} \int_{0}^{\infty} \frac{x}{(x^{2} + 1)} dx + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{1 + x^{2}}$

$= - \frac{1}{2} {[\log (1 + x)]}_{0}^{\infty} + \frac{1}{4} {[\log (x^{2} + 1)]}_{0}^{\infty} + \frac{1}{2} \times \frac{π}{2} = - \frac{1}{2} \lim_{x \to \infty} \log (1 + x) + \frac{1}{4} \lim_{x \to \infty} \log (1 + x^{2}) + \frac{π}{4} = \lim_{x \to \infty} \log [\frac{{(1 + x^{2})}^{\frac{1}{4}}}{{(1 + x)}^{\frac{1}{2}}}] + \frac{π}{4} = \lim_{x \to \infty} \log [\frac{\sqrt{x} {(\frac{1}{x^{2}} + 1)}^{\frac{1}{4}}}{\sqrt{x} {(\frac{1}{x} + 1)}^{\frac{1}{2}}}] + \frac{π}{4} = \log (\frac{{(0 + 1)}^{\frac{1}{4}}}{{(0 + 1)}^{\frac{1}{2}}}) + \frac{π}{4} = \log (1) + \frac{π}{4} = 0 + \frac{π}{4} = \frac{π}{4}$